To simplify the expression:

$\frac{1}{3^{5n}} \times 9^{n-1} \times 27^{n+1}$

Step-by-step solution:

1. Rewrite the terms using powers of 3:

   • $9 = 3^2$, so $9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2}$.
   • $27 = 3^3$, so $27^{n+1} = (3^3)^{n+1} = 3^{3(n+1)} = 3^{3n+3}$.

   Now the expression becomes:

   $\frac{1}{3^{5n}} \times 3^{2n-2} \times 3^{3n+3}$

2. Combine the powers of 3:

   Using the rule $a^m \times a^n = a^{m+n}$, we can combine the exponents of 3.

   $3^{-(5n)} \times 3^{2n-2} \times 3^{3n+3} = 3^{-(5n) + (2n-2) + (3n+3)}$

3. Simplify the exponent:

   Combine the terms in the exponent:

   $-(5n) + (2n-2) + (3n+3) = -5n + 2n + 3n - 2 + 3 = 0 + 1 = 1$

   So, the expression simplifies to:

   $3^1 = 3$

Final Answer:

The simplified expression is:

$3$

Would you like more details, or do you have any questions?

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Here are 5 related questions for further practice:

1. Simplify $\frac{1}{2^{3n}} \times 4^{n-1} \times 16^{n+2}$.
2. Simplify $\frac{1}{5^{4n}} \times 25^{n} \times 125^{n-1}$.
3. Express $8^{2n} \times \frac{1}{2^{6n}}$ in terms of powers of 2.
4. Simplify $2^{3x-2} \times 4^{x+1} \times \frac{1}{8^x}$.
5. If $x^a \times x^b = x^c$, what is $c$ in terms of $a$ and $b$?

Tip: Remember that when multiplying terms with the same base, you add their exponents.